Question

A thin film of water $(\mu \, = \dfrac{4}{3})$ is$3100{\rm{\dot A}}$. If it is illuminated by white light at normal incidence, the colour of the film in the reflected light will be:
A. Blue
B. Black
C. Yellow
D. Red

Answer 1

Optical properties of thin films can be obtained by interference and reflection. When a white light incident occurs on a thin film, the film appears coloured and the colour depends on the thickness of the film and the angle of the incident of light. The thin films between two media obtained the interference pattern, i.e., constructive and destructive interference of different wavelengths.

Complete step by step answer:
Given data,
Refractive index of water $\mu = \;\dfrac{4}{3}$
The thickness of the film is $= \,3100\;\mathop {\rm{A}}\limits^{\rm{o}} \, = \,310 \times {10^{ - 9}}\;{\rm{m}}$
Here the thin film forms constructive interference, So we apply the constructive interference formula to find wavelengths.
Formula used:
$\Delta {\rm{x = }}\mu {\rm{t = }}\left( {{\rm{m + }}\dfrac{1}{2}} \right)\dfrac{\lambda }{2}$
The path difference is $\Delta {\rm{x}}$, $\mu$ is the refractive index,$\lambda$ which is the wavelength, and t is the film's thickness.
$\Rightarrow \lambda \; = \,\dfrac{{2\mu {\rm{t}}}}{{\left( {{\rm{m + }}\dfrac{1}{2}} \right)}}$
Now substitute the given value in this formula we get,
$\Rightarrow {\lambda _0}\; = \;\dfrac{{2 \times 1.33 \times 3100 \times {{10}^{ - 9}}}}{{\left( {0 + \dfrac{1}{2}} \right)}}$ (Where m =0, 1, 2, 3…………)
$\Rightarrow {\lambda _0}\; = \dfrac{{8246 \times {{10}^{ - 9}}}}{{\dfrac{1}{2}}}\, = \;\dfrac{{8246 \times {{10}^{ - 9}} \times 2}}{1} = 1649nm$ (Infrared region)
Similarly, ${\lambda _1}\; = \;\dfrac{{2 \times 1.33 \times 3100 \times {{10}^{ - 9}}}}{{\left( {1 + \dfrac{1}{2}} \right)}}$
$\Rightarrow {\lambda _1}\; = \dfrac{{8246 \times {{10}^{ - 9}}}}{{\dfrac{3}{2}}}\, = \;\dfrac{{8246 \times {{10}^{ - 9}} \times 2}}{3} = 549nm$

${\lambda _2}\; = \;\dfrac{{2 \times 1.33 \times 3100 \times {{10}^{ - 9}}}}{{\left( {2 + \dfrac{1}{2}} \right)}}$
${\lambda _2}\; = \dfrac{{8246 \times {{10}^{ - 9}}}}{{\dfrac{5}{2}}}\, = \;\dfrac{{8246 \times {{10}^{ - 9}} \times 2}}{5} = 330nm$ (Ultraviolet region)
Thus 1649nm and 330nm wavelength correspond to light, which is visible to the human eye. Therefore reflection occurs at 549nm, which leads to the blue-violet region of the spectrum. Suppose reflected light is weak in the red region of the spectrum and strong in the blue-violet region. So the film possesses a blue colour.

Therefore the correct option is (A).

Note: Constructive interference can only occur when two waves are in phase and having the same frequency, constant phase difference. If the waves are in our phase and have the same or different frequencies and no constant phase difference, then the destructive interference occurs.